In the circuit shown in the figure,the potent | vedclass

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(D) The circuit consists of two capacitors with capacitances $C_1 = 4.5\,\mu F$ (which is $\frac{9}{2}\,\mu F$) and $C_2 = 9\,\mu F$ connected in seri

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$8$

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(D) The circuit consists of two capacitors with capacitances $C_1 = 4.5\,\mu F$ (which is $\frac{9}{2}\,\mu F$) and $C_2 = 9\,\mu F$ connected in series with a $12\,V$ $DC$ source.In a series combination of capacitors,the charge $Q$ on each capacitor is the same.The equivalent capacitance $C_{eq}$ is given by:$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{4.5} + \frac{1}{9} = \frac{2}{9} + \frac{1}{9} = \frac{3}{9} = \frac{1}{3}\,\mu F^{-1}$So,$C_{eq} = 3\,\mu F$.The total charge $Q$ supplied by the battery is:$Q = C_{eq} 	imes V_{total} = 3\,\mu F 	imes 12\,V = 36\,\mu C$.The potential difference $V_1$ across the $4.5\,\mu F$ capacitor is:$V_1 = \frac{Q}{C_1} = \frac{36\,\mu C}{4.5\,\mu F} = 8\,V$.

In the circuit shown in the figure,the potential difference across the $4.5\,\mu F$ capacitor is ....... $V$.

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